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Question I was given: Let V be a uniform random variable distributed over the interval (0,1). Let $\ X = \frac{1}{\sqrt(U)}$. What is the cumulative distribution function and probability density function of X?

I understand the basic concept here behind cumulative distribution functions and probability density functions. What's throwing me off here is the additional parameter saying that V is a uniform random variable.

Is the cumulative distance function just: $\ \int\limits_0^1 \frac{1}{\sqrt(V)}dV $ which would then just equal 2?

Then the prob dens function would be: $$ \frac{1}{\sqrt(V)}-----0<V<1 $$ $$ 0------otherwise $$ This seems too simple to me to be the correct answer though. Am I missing something? Am I finding the cumulative distance function of V rather than of X?

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up vote 0 down vote accepted

We find the cumulative distribution function $F_X(x)$ of $X$. So we want to find $\Pr(X\le x)$.

If $x\le 1$, then $F_X(x)=0$. For since $0\lt U\lt 1$, it is impossible for $\dfrac{1}{\sqrt{U}}$ to be $\le 1$.

Now we find $F_X(x)$ for $x\gt 1$.

We have $X\le x$ if and only if $\dfrac{1}{\sqrt{U}}\le x$. Flip it over. So $X\le x$ if and omly if $\sqrt{U}\ge \dfrac{1}{x}$, that is, if and only if $U\ge \dfrac{1}{x^2}$.

But $\Pr\left(U\ge \dfrac{1}{x^2}\right)=1-\Pr\left(U\le \dfrac{1}{x^2}\right)$. Since $U$ is uniform, this is simply $1-\dfrac{1}{x^2}$. Thus for $X\gt 1$, $$F_X(x)= 1-\frac{1}{x^2}.$$

For the density function $f_X(x)$, differentiate the cdf.

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Which would just yield 2/X^3 for X>1 and 0 otherwise? – chrisbster Mar 21 '13 at 2:24
And thank you - that clarification was fantastic! – chrisbster Mar 21 '13 at 2:24
You are welcome. Yes, the density function (for $x\gt 1$) is $\frac{2}{x^3}$. – André Nicolas Mar 21 '13 at 2:32

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